A car starts from rest and moves with uniform acceleration. If it covers 5.5 m in the 4th second and 9.5 m in the 8th second, find its acceleration and initial velocity.

The distance $S_n$ covered by an object in the $n$-th second of its motion, starting with initial velocity $u$ and moving with uniform acceleration $a$, is given by:

$$S_n = u + \frac{a}{2}(2n - 1)$$

For the 4th second, $n=4$ and $S_4 = 5.5$ m:

$$5.5 = u + \frac{a}{2}(2 \times 4 - 1)$$ $$5.5 = u + \frac{a}{2}(8 - 1)$$ $$5.5 = u + \frac{7a}{2} \quad \text{(Equation 1)}$$

For the 8th second, $n=8$ and $S_8 = 9.5$ m:

$$9.5 = u + \frac{a}{2}(2 \times 8 - 1)$$ $$9.5 = u + \frac{a}{2}(16 - 1)$$ $$9.5 = u + \frac{15a}{2} \quad \text{(Equation 2)}$$

Subtract Equation 1 from Equation 2:

$$\left(u + \frac{15a}{2}\right) - \left(u + \frac{7a}{2}\right) = 9.5 - 5.5$$ $$\frac{15a}{2} - \frac{7a}{2} = 4.0$$ $$\frac{8a}{2} = 4.0$$ $$4a = 4.0$$ $$a = \frac{4.0}{4}$$ $$a = 1.0\, \text{m/s}^2$$

$$5.5 = u + \frac{7(1.0)}{2}$$ $$5.5 = u + \frac{7}{2}$$ $$5.5 = u + 3.5$$ $$u = 5.5 - 3.5$$ $$u = 2.0\, \text{m/s}$$

The acceleration of the car is $\boxed{\text{1.0 m/s}^2}$ and its initial velocity is $\boxed{\text{2.0 m/s}}$.